海亮 7.21 模拟赛

海亮 7.21 模拟赛

累了 随便打打 不切题就不切题罢

#A. 手语的 (gnsi)

只需要判断三倍重心点的$x,y$坐标是否在三个矩形的$x,y$坐标最小值和最大值加和之内即可

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long 
int read ()
{
	int x = 0 , f = 1;
	char ch = cin.get();
	while ( !isdigit ( ch ) ) { if ( ch == '-' ) f = -1; ch = cin.get(); }
	while ( isdigit ( ch ) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = cin.get(); }
	return x * f;
}

int xl[3] , xr[3] , yl[3] , yr[3];

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	int T = read();
	while ( T -- )
	{
		for ( int i = 0 ; i < 3 ; i ++ ) xl[i] = read() , xr[i] = read() , yl[i] = read() , yr[i] = read();
		int x = read() , y = read();
		cout << ( ( xl[0] + xl[1] + xl[2] <= 3 * x && xr[0] + xr[1] + xr[2] >= 3 * x && yl[0] + yl[1] + yl[2] <= 3 * y && yr[0] + yr[1] + yr[2] >= 3 * y ) ? "heihei" : "yiyandingzhen" ) << endl;
	}
	return 0;
}

#B. 序列的 (quencese)

设置$d{p}_j$表示以数字$j$为结尾的数列可能的状态

那么显然有$d{p}_i=\sum _{j=1}^{\lfloor i\rfloor }d{p}_j$

那么我们可以看到答案是一段前缀和 记录一段前缀和就足够过$50$的点 但还不够

考虑继续优化 削减掉第二次操作 那么我们可以预处理$max{n}^{\frac{1}{4}}$以内的所有数组

那么对于$i$ 它会对$i^2$及以后的$dp$值产生贡献

那么$d{p}_i$对答案的贡献就是$d{p}_i\ast (sqrt(x)-i^2+1)$

累加即可 时间复杂度$O(max{n}^{\frac{1}{4}})$

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define pb push_back
#define int __int128
constexpr int mod = 998244353;
constexpr int N = 1e6 + 5;
char buf[1<<22] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin))?EOF:*p1++)

int read ()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit ( ch ) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit ( ch ) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , x , f[N];

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	f[1] = 1;
	for ( int i = 2 ; i <= 100000 ; i ++ )
		for ( int j = 1 ; j <= (int)sqrt((long double)i) ; j ++ ) f[i] += f[j];
	int T = read();
	while ( T -- )
	{
		x = read();
		int xx = sqrt((long double)x);
		int xxx = sqrt((long double)xx);
		int ans = 0;
		for ( int i = 1 ; i <= xxx ; i ++ ) 
			ans += f[i] * ( xx - i * i + 1 );
		cout << (long long)ans << endl;
	}
	return 0;
}

#D. 好的 (doog)

我的评价是:暴力出奇迹

$50pts$代码

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define pb push_back
constexpr int N = 5e5 + 5;
int read ()
{
	int x = 0 , f = 1;
	char ch = cin.get();
	while ( !isdigit ( ch ) ) { if ( ch == '-' ) f = -1; ch = cin.get(); }
	while ( isdigit ( ch ) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = cin.get(); }
	return x * f;
}

int n , q , a[N] , buc[N];

string s;

void add ( int l , int r )
{
	for ( int i = l ; i <= r ; i ++ ) a[i] = ( a[i] + 1 ) % 3;
}

int query ( int l , int r )
{
	int pd = 1;
	buc[0] = 0 , buc[1] = 0 , buc[2] = 0;
	for ( int i = l ; i <= r ; i ++ ) buc[a[i]] ++;
	if ( ( buc[0] & 1 ) || ( buc[1] & 1 ) || ( buc[2] & 1 ) ) pd = 0;
	if ( pd == 0 ) return 0;
	vector<int> vec;
	for ( int i = l ; i <= r ; i ++ ) vec.push_back(a[i]);
	while ( !vec.empty() )
	{
		bool jian = 1;
		for ( int i = 0 ; i < (int)vec.size() - 1 ; i ++ ) if ( vec[i] == vec[i+1] ) vec.erase ( vec.begin() + i , vec.begin() + i + 2 ) , jian = 0;
		if ( jian ) { pd = 0; break; }
	}
	return pd;
}
/*
8 9
01211012
2 4 5
2 3 6
1 6 8
1 6 8
2 3 6
2 1 8
1 1 1
1 7 7
2 1 8
*/

signed main ()
{
//	freopen ( "doog1.in" , "r" , stdin );
//	freopen ( "doog1.out" , "w" , stdout );
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , q = read();
	cin >> s;
	for ( int i = 1 ; i <= n ; i ++ ) a[i] = s[i-1] - '0';
	while ( q -- )
	{
		int op = read() , l = read() , r = read();
		if ( op == 1 ) add ( l , r );
		else cout << ( query ( l , r ) ? "YES" : "NO" ) << endl;
	}
	return 0;
}